no, it cannot

if n/2*(2*n+2111)=m*m m is an integer

then 2*n^2+2111*n-2*m^2=0

and 2111+16m^2=k^2 k is an integer

or (k-t )*(k+t )=2111 t=4*m

and since 2111 is a prime number

there is only one integer solution k=1056 t=1055

but t=4*m so m cannot be an integer

qed